Finite sum of eigenspaces (with distinct eigenvalues) is a direct sum

No, this is not a full proof. It is not true that, if $V = A+B+C$, and $A \cap B = A \cap C = B \cap C = \{ 0 \}$, then $V = A \oplus B \oplus C$. For example, let $V = \mathbb{C}^2$ and let $A$, $B$ and $C$ be the one dimensional subspaces spanned by $(1,0)$, $(1,1)$ and $(0,1)$.

This does give some good intuition for why the claim is true. If you want to build your way to the full proof, you might try the special case of three eigenspaces and see what you can do.

Amusingly, this is currently the top voted example of a common false belief over at MO.

To clarify some concepts, I list some relevant definitions from Artin's "Algebra". Suppose $\{W_i\}$ $i=1,\ldots,n$ are vector subspaces of $V$, then their sum is given by $$ W_1 + \cdots + W_n := \{v\in V\ |\ v = w_1 + \cdots + w_n \mbox{ ,with } w_i\in W_i\} $$ The subspaces $W_1,\ldots,W_n$ are independent if $$ w_1 + \cdots + w_n = 0 \mbox{ ,with } w_i\in W_i \mbox{ implies } w_i = 0 \mbox{ for all i} $$ A subspace $U$ is called a direct sum of $W_1,\ldots,W_n$ if $U = W_1 + \cdots + W_n$ and $W_1,\ldots,W_n$ are independent.

From the above definitions, what we real need to show in this problem is that eigenspaces corresponding to distinct eigenvalues are independent.

Let $v_1,\ldots,v_n$ be eigenvectors with eigenvalues $\lambda_1,\ldots,\lambda_n$, respectively, and $v_1 + \cdots + v_n = 0$.

Now, we use induction to show all $v_i = 0$. If $n = 1$, it's trivial. Otherwise, we have the following two observations. If we multiply both sides of the equation by $\lambda_1$, we get $$ \lambda_1 v_1 + \cdots + \lambda_1 v_n = 0 $$ If we apply the linear map $\alpha$ to both sides of the equation, we get $$ \lambda_1 v_1 + \cdots + \lambda_n v_n = 0 $$ Then, we subtract the first equation from the second one $$ (\lambda_2 - \lambda_1) v_2 + \cdots + (\lambda_n - \lambda_1) v_n = 0 $$ Since $\lambda_1,\ldots,\lambda_n$ are distinct, the coefficients in the equation above are non-zero.

By induction $v_1 = \cdots = v_n = 0$.

Hence, eigenspaces corresponding to distinct eigenvalues are independent.